Consider the LPP: Maximize \(z = 5x + 3y\) subject to \(3x + 5y \le 15, 5x + 2y \le 10, x, y \ge 0\). The optimal feasible solution occurs at
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left(\displaystyle\frac{20}{19},\frac{45}{19}\right)\) only
The feasible region is bounded, so the maximum of z occurs at a corner point.
\(3x + 5y = 15\) meets the axes at (5, 0) and (0, 3).
\(5x + 2y = 10\) meets the axes at (2, 0) and (0, 5).
The two lines meet where
\(15x + 25y = 75\)
\(15x + 6y = 30\)
Subtracting:
\(\displaystyle 19y = 45,\quad y = \frac{45}{19}\)
\(\displaystyle x = \frac{15 - 5y}{3}\)
\(\displaystyle = \frac{1}{3}\left(15 - \frac{225}{19}\right) = \frac{20}{19}\)
Both constraints are of the “≤” type, so the region lies on the origin side of both lines, in the first quadrant. Its corners are (0, 0), (2, 0), \(\left(\frac{20}{19}, \frac{45}{19}\right)\) and (0, 3).
\(z(0, 0) = 0\)
\(z(2, 0) = 10\)
\(z(0, 3) = 9\)
\(\displaystyle z\left(\frac{20}{19}, \frac{45}{19}\right) = \frac{100 + 135}{19} = \frac{235}{19} \approx 12.37\)
The largest value, \(\frac{235}{19}\), occurs only at \(\left(\frac{20}{19}, \frac{45}{19}\right)\).
NCERT (2026-27 edition):
Graphical Method of Solving Linear Programming Problems
